3.74 \(\int \frac {1}{a+b \text {csch}(c+d x)} \, dx\)

Optimal. Leaf size=54 \[ \frac {2 b \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}+\frac {x}{a} \]

[Out]

x/a+2*b*arctanh((a-b*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/a/d/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3783, 2660, 618, 204} \[ \frac {2 b \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}+\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x])^(-1),x]

[Out]

x/a + (2*b*ArcTanh[(a - b*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a*Sqrt[a^2 + b^2]*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{a+b \text {csch}(c+d x)} \, dx &=\frac {x}{a}-\frac {\int \frac {1}{1+\frac {a \sinh (c+d x)}{b}} \, dx}{a}\\ &=\frac {x}{a}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {1}{1-\frac {2 i a x}{b}+x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a d}\\ &=\frac {x}{a}-\frac {(4 i) \operatorname {Subst}\left (\int \frac {1}{-4 \left (1+\frac {a^2}{b^2}\right )-x^2} \, dx,x,-\frac {2 i a}{b}+2 \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a d}\\ &=\frac {x}{a}+\frac {2 b \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}-\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 64, normalized size = 1.19 \[ \frac {-\frac {2 b \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{d \sqrt {-a^2-b^2}}+\frac {c}{d}+x}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x])^(-1),x]

[Out]

(c/d + x - (2*b*ArcTan[(a - b*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/(Sqrt[-a^2 - b^2]*d))/a

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fricas [B]  time = 0.41, size = 186, normalized size = 3.44 \[ \frac {{\left (a^{2} + b^{2}\right )} d x + \sqrt {a^{2} + b^{2}} b \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b\right )}}{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) + 2 \, {\left (a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) - a}\right )}{{\left (a^{3} + a b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c)),x, algorithm="fricas")

[Out]

((a^2 + b^2)*d*x + sqrt(a^2 + b^2)*b*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + a^
2 + 2*b^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(a*cosh(d*x + c) + a*sinh(d*x + c) +
b))/(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)*sinh(d*x + c) - a)))
/((a^3 + a*b^2)*d)

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giac [A]  time = 0.14, size = 84, normalized size = 1.56 \[ -\frac {\frac {b \log \left (\frac {{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a} - \frac {d x + c}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c)),x, algorithm="giac")

[Out]

-(b*log(abs(2*a*e^(d*x + c) + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^(d*x + c) + 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a
^2 + b^2)*a) - (d*x + c)/a)/d

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maple [A]  time = 0.20, size = 87, normalized size = 1.61 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}-\frac {2 b \arctanh \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d a \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*csch(d*x+c)),x)

[Out]

-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-2/d/a*b/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1
/2*d*x+1/2*c)*b-2*a)/(a^2+b^2)^(1/2))

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maxima [A]  time = 0.41, size = 85, normalized size = 1.57 \[ -\frac {b \log \left (\frac {a e^{\left (-d x - c\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-d x - c\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {d x + c}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c)),x, algorithm="maxima")

[Out]

-b*log((a*e^(-d*x - c) - b - sqrt(a^2 + b^2))/(a*e^(-d*x - c) - b + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a*d) +
(d*x + c)/(a*d)

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mupad [B]  time = 0.33, size = 121, normalized size = 2.24 \[ \frac {x}{a}-\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{c+d\,x}\right )}{a^2\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}}+\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2}+\frac {2\,b\,\left (a-b\,{\mathrm {e}}^{c+d\,x}\right )}{a^2\,\sqrt {a^2+b^2}}\right )}{a\,d\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/sinh(c + d*x)),x)

[Out]

x/a - (b*log((2*b*exp(c + d*x))/a^2 - (2*b*(a - b*exp(c + d*x)))/(a^2*(a^2 + b^2)^(1/2))))/(a*d*(a^2 + b^2)^(1
/2)) + (b*log((2*b*exp(c + d*x))/a^2 + (2*b*(a - b*exp(c + d*x)))/(a^2*(a^2 + b^2)^(1/2))))/(a*d*(a^2 + b^2)^(
1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \operatorname {csch}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c)),x)

[Out]

Integral(1/(a + b*csch(c + d*x)), x)

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